Can we expect $\|fg\|_{\mathcal{F}L^{1}} \leq C \|f\|_{L^{2}(\mathbb R)} \|g\|_{\mathcal{F}L^{1}}$?

Question

Consider the space of all Fourier transforms of $L^{1}(\mathbb R),$ that is, $$\mathcal{F}L^{1}=\mathcal{F}L^{1}(\mathbb R):= \{f\in L^{\infty}(\mathbb R):\hat{f}\in L^{1}(\mathbb R)\},$$ with the norm, $\|f\|_{\mathcal{F}(L^{1})}:=\|\hat{f}\|_{L^{1}(\mathbb R)}.$ We note that the space $\mathcal{F}L^{1}$ is an algebra under poinwise multiplication.

My Question:Can we expect: $\|fg\|_{\mathcal{F}L^{1}} \leq C \|f\|_{L^{2}(\mathbb R)} \|g\|_{\mathcal{F}L^{1}}$? (where $C$ is some constant)

Thanks,

Answer 1

Since your question is about this inequality \begin{align*} \|fg\|_{\mathcal{F}L^{1}} \leq C \|f\|_{L^{2}} ~\|g\|_{\mathcal{F}L^{1}}~, \tag{1} \end{align*} the preconditions should be $f\in L^2$ and $g \in \mathcal{F}L^{1}$. However, using only these two preconditions could hardly lead to such a result. On the other hand, by adding some other conditions, we will find inequality $(1)$ is valid ( not always ). Next, I will prove the following statement:

Suppose $f\in L^2$, $g \in L^2 \bigcap \mathcal{F}L^{1}$ and $\widehat{fg}(x) = \mathcal{O}\left( |x|^{-3} \right)$, as $|x| \to \infty$. Then we have $fg \in \mathcal{F}L^{1}$ and either inequality $(1)$ is valid or $\|fg\|_{\mathcal{F}L^{1}} \leq C \|f\|^{\frac{1}{2}}_{L^{2}} ~\|g\|^{\frac{1}{2}}_{\mathcal{F}L^{1}}$.

Proof: Since $f,g \in L^2$, we have $\widehat{fg} = \hat{f} * \hat{g}$. Hence, $$\| \widehat{fg} \|_{L^2} = \| \hat{f} * \hat{g} \|_{L^2} \leq \| \hat{f} \|_{L^2} \| \hat{g} \|_{L^1} = \| f \|_{L^2} \| g \|_{\mathcal{F}L^{1}}~~.$$ To estimate $\| \widehat{fg} \|_{L^1}$, we have \begin{align*} \int_{\mathbb{R}} |\widehat{fg}|dx &\leq \left( \int_{\mathbb{R}} \frac{1}{1+x^2}dx \right)^{\frac{1}{2}} \left( \int_{\mathbb{R}} \left( 1 + x^2 \right) |\widehat{fg}(x)|^2 dx \right)^{\frac{1}{2}} \\ &\leq \sqrt{\pi} \left( \| \widehat{fg} \|_{L^2} + \| x \widehat{fg} \|_{L^2} \right). \tag{2} \end{align*} In order to estimate $\| x \widehat{fg} \|_{L^2}$, we do it piecewise. Since $\widehat{fg}(x) = \mathcal{O}\left( |x|^{-3} \right)$ as $|x| \to \infty$, there are constants $C_0>0$ and $x_0 >1$ such that $|x \widehat{fg}(x)| \leq \frac{C_0}{|x|^2}$ when $|x|>x_0$. Then we get \begin{align*} \int_{\mathbb{R}} |x \widehat{fg}(x)|^2 dx &= \int_{|x|<x_0} |x \widehat{fg}(x)|^2 dx + \int_{|x|\geq x_0} |x \widehat{fg}(x)|^2 dx \\ &\leq x^2_0 \int_{\mathbb{R}} |\widehat{fg}(x)|^2 dx + C_0 \int_{|x| \geq x_0} \frac{|\widehat{fg}(x)|}{|x|} dx \\ &\leq x^2_0 \int_{\mathbb{R}} |\widehat{fg}(x)|^2 dx + C_0 \left( \int_{|x|\geq x_0} \frac{1}{x^2} dx \right)^{\frac{1}{2}} \left( \int_{\mathbb{R}} |\widehat{fg}(x)|^2 dx \right)^{\frac{1}{2}} \\ &\leq C_1 H, \end{align*} where $H = \max \{ \| \widehat{fg} \|^2_{L^2}~, ~ \| \widehat{fg} \|_{L^2} \}$. Therefore, combining $(2)$ we have $$\| \widehat{fg} \|_{L^1} \leq CH^{\frac{1}{2}}.$$ This completes the proof. $~\square$

From this proof, we find that inequality $(1)$ is valid if $\| \widehat{fg} \|_{L^2} \geq 1$.

Corollary $~$ Suppose $f,~g \in L^1 \bigcap \mathcal{F}L^{1}$ and $\widehat{fg}(x) = \mathcal{O}\left( |x|^{-3} \right)$, as $|x| \to \infty$. Then we have that either inequality $(1)$ is valid or $\|fg\|_{\mathcal{F}L^{1}} \leq C \|f\|^{\frac{1}{2}}_{L^{2}} ~\|g\|^{\frac{1}{2}}_{\mathcal{F}L^{1}}$.